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2 years ago

Use linear Approximation to approximate e^.01

1 answer:

5 0

Suppose that:
let f(x) = e^x

Suppose:
a = 0
Now we are finding f(a):
f(a) = f(0) = 1

f '(x) = e^x
Now we are finding f '(a):
f '(a) = f '(0) = 1
Putting the formula:
L(x) = f(a) + f '(a)( x - a)

L(e^0.01) = 1 + 1(0.01 - 0)

= 1.01 is the answer

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Find the midpoint of the line joining A (3,5) and B(1,3)

MaRussiya [10]

Answer:

(-1 , -1)

Step-by-step explanation:

$x_1= 3 \\ y_1= 5 \\ x_2=1 \\ y_2= 3$

$midpoint \: = ( \frac{x_2 - x_1}{2} \: \frac{y_2 -y_1}{2} )$

$=( \frac{1 - 3}{2} \: \frac{3 - 5}{2} ) \\ = ( \frac{ - 2}{2} \: \frac{ - 2}{2} ) = \\ ( - 1 \: - 1)$

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3 years ago

Divide. Write your answer in simplest form. −8/9÷(−8/9)=

MakcuM [25]

Answer:

$- \frac{8}{9} \div ( - \frac{8}{9} ) \\ - \frac{8}{9} \times ( - \frac{9}{8} ) \\ = 1$

I hope I helped you^_^

5 0

2 years ago

Pls help will give you good rating

son4ous [18]

Answer:

-1

Step-by-step explanation:

3 0

2 years ago

Find the range of each function for the domain {-2,0,3.5}. g(x)=3x-5

yaroslaw [1]

Hello from MrBillDoesMath!

Answer:

Range = { -11, -5, 4, 10}

Discussion:

g(x) = 3x- 5 =>

g(-2) = 3(-2) -5 = -6 -5 = -11

g(0) = 3(0) - 5 = 0 -5 = -5

g(3) = 4(3)-5 = 9-5 =4

g(5) = 3(5)-5 =15-5 = 10

So the range is { -11, -5, 4, 10}

Thank you,

MrB

7 0

2 years ago

Can someone please please help me with this???

Sholpan [36]

9514 1404 393

Answer:

25 terms

Step-by-step explanation:

We can use these numbers to find the first term and the common difference. Then we can use those in the formula for the sum of terms. Finally, we can solve the inequality that makes the sum of terms greater than 1560.

First term and common difference

The general term is ...

an = a1+d(n -1)

Then the given terms are ...

a2 = a1 + d(2 -1) . . . second term

8 = a1 + d . . . . . . . . . [eq1]

a4 = a1 + d(4 -1) . . . fourth term

18 = a1 + 3d . . . . . . . [eq2]

Subtracting the first equation from the second, we get ...

[eq2] -[eq1] = (18) -(8) = (a1+3d) -(a1 +d)

10 = 2d . . . . . simplify

5 = d . . . . . . . divide by 2

Using [eq1], we can find a1:

8 = a1 + 5

3 = a1 . . . . . . subtract 5

The arithmetic progression has a first term of 3 and a common difference of 5.

Formula for the sum of terms

The generic formula for the sum of an arithmetic progression is ...

Sn = n·(a1 + (a1 +d(n -1)))/2

Sn = (n/2)(2a1 + d(n -1)) . . . . simplify a little bit

With our values for a1 and d, this becomes ...

Sn = (n/2)(2·3 +5(n -1))

Sn = (n/2)(5n +1)

Inequality for the sum

We want the sum of terms to be more than 1560, so we can write the inequality ...

(n/2)(5n +1) > 1560

In standard form, this would be ...

5n² +n -3120 > 0

The quadratic formula tells us the zeros are ...

n = (-1 ±√(1² -4(5)(-3210)))/(2·5) = (-1 ± √62401)/10

The positive zero is n ≈ 24.88.

That is, for n > 24.9, the sum will be greater than 1560.

The least number of terms for a sum greater than 1560 is 25.

Check

Check the formula for 4 terms.

3 + 8 + 13 + 18 = 42 = (4/2)(5·4 +1) = 2(21) = 42 . . . note 2nd & 4th terms

Check the sums for 24 and 25 terms.

(24/2)(5·24 +1) = 12(121) = 1452 . . . . . sum < 1560 for n = 24

(25/2)(5·25 +1) = 12.5(126) = 1575 . . . sum > 1560 for n = 25

4 0

2 years ago